C Program to solve the producer consumer problem

Write a C Program to solve the producer consumer problem with two processes using semaphores.
Producer-consumer problem is the standard example of multiple process synchronization problem. The problem occurs when concurrently producer and consumer tries to fill the data and pick the data when it is full or empty. producer consumer problem is also known as bounded-buffer problem. In this program We use the semaphores, to solve the problem.Read more about C Programming Language . and read the C Programming Language (2nd Edition) by K and R.

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#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <time.h>
#include <sys/types.h>
#include <sys/ipc.h>
#include <sys/sem.h>


#define NUM_LOOPS 20
int main(int argc, char* argv[])
{
int sem_set_id;
union semun sem_val;
int child_pid;
int i;
struct sembuf sem_op;
int rc;
struct timespec delay;


sem_set_id = semget(IPC_PRIVATE, 1, 0600);
if (sem_set_id == -1) {
perror("main: semget");
exit(1);
}
printf("Semaphore set created,
semaphore set id '%d'.n", sem_set_id);


sem_val.val = 0;
rc = semctl(sem_set_id, 0, SETVAL, sem_val);
child_pid = fork();
switch (child_pid) {
case -1:
perror("fork");
exit(1);
case 0:
for (i=0; i<NUM_LOOPS; i++) {
sem_op.sem_num = 0;
sem_op.sem_op = -1;
sem_op.sem_flg = 0;
semop(sem_set_id, &sem_op, 1);
printf("consumer: '%d'n", i);
fflush(stdout);
sleep(3);
}
break;
default:
for (i=0; i<NUM_LOOPS; i++)
{
printf("producer: '%d'n", i);
fflush(stdout);
sem_op.sem_num = 0;
sem_op.sem_op = 1;
sem_op.sem_flg = 0;
semop(sem_set_id, &sem_op, 1);
sleep(2);
if (rand() > 3*(RAND_MAX/4))
{
delay.tv_sec = 0;
delay.tv_nsec = 10;
nanosleep(&delay, NULL);
}
}
break;
}


return 0;
}
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C Program to solve Dining philosophers problem.

Write a C Program to solve Dining philosophers problem.
Dining philosophers problem is a classic synchronization problem.A problem introduced by Dijkstra concerning resource allocation between processes. Five silent philosophers sit  around table with a bowl of spaghetti. A fork is placed between each pair of adjacent philosophers.

Each philosopher must alternately think and eat.
Eating is not limited by the amount of spaghetti left: assume an infinite supply.
However, a philosopher can only eat while holding both the fork to the left and the fork to the right
(an alternative problem formulation uses rice and chopsticks instead of spaghetti and forks).

Each philosopher can pick up an adjacent fork, when available, and put it down, when holding it.
These are separate actions: forks must be picked up and put down one by one.
The problem is how to design a discipline of behavior (a concurrent algorithm) such that each philosopher won’t starve, i.e. can forever continue to alternate between eating and thinking.
Read more about C Programming Language.

/***********************************************************
* You can use all the programs on www.c-program-example.com
* for personal and learning purposes. For permissions to use the
* programs for commercial purposes,
* contact info@c-program-example.com
* To find more C programs, do visit www.c-program-example.com
* and browse!
*
* Happy Coding
***********************************************************/

#include<stdio.h>
#include<semaphore.h>
#include<pthread.h>

#define N 5
#define THINKING 0
#define HUNGRY 1
#define EATING 2
#define LEFT (ph_num+4)%N
#define RIGHT (ph_num+1)%N

sem_t mutex;
sem_t S[N];

void * philospher(void *num);
void take_fork(int);
void put_fork(int);
void test(int);

int state[N];
int phil_num[N]={0,1,2,3,4};

int main()
{
int i;
pthread_t thread_id[N];
sem_init(&mutex,0,1);
for(i=0;i<N;i++)
sem_init(&S[i],0,0);
for(i=0;i<N;i++)
{
pthread_create(&thread_id[i],NULL,philospher,&phil_num[i]);
printf("Philosopher %d is thinkingn",i+1);
}
for(i=0;i<N;i++)
pthread_join(thread_id[i],NULL);
}

void *philospher(void *num)
{
while(1)
{
int *i = num;
sleep(1);
take_fork(*i);
sleep(0);
put_fork(*i);
}
}

void take_fork(int ph_num)
{
sem_wait(&mutex);
state[ph_num] = HUNGRY;
printf("Philosopher %d is Hungryn",ph_num+1);
test(ph_num);
sem_post(&mutex);
sem_wait(&S[ph_num]);
sleep(1);
}

void test(int ph_num)
{
if (state[ph_num] == HUNGRY && state[LEFT] != EATING && state[RIGHT] != EATING)
{
state[ph_num] = EATING;
sleep(2);
printf("Philosopher %d takes fork %d and %dn",ph_num+1,LEFT+1,ph_num+1);
printf("Philosopher %d is Eatingn",ph_num+1);
sem_post(&S[ph_num]);
}
}

void put_fork(int ph_num)
{
sem_wait(&mutex);
state[ph_num] = THINKING;
printf("Philosopher %d putting fork %d and %d downn",ph_num+1,LEFT+1,ph_num+1);
printf("Philosopher %d is thinkingn",ph_num+1);
test(LEFT);
test(RIGHT);
sem_post(&mutex);
}
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